Find points on the curve `(x^2)/9+(y^2)/(16)=1` at which the tangents are parallel to the y-axis.

Options

A. (0,4) & (0,-4)
B. (4,0) & (-4,0)
C. (3,0) & (-3,0)
D. (0,3) & (0,-3)

Answer

Option C

Points are `(3,0)` and `(-3,0)`.

Explanation

The equation of the given curve is `\frac{x^{2}}{9}+\frac{y^{2}}{16}=1`


On differentiating both sides wrt `x`, we have:

$$ \begin{aligned} &\frac{2 x}{9}+\frac{2 y}{16} \frac{d y}{d x}=0 \\ &\Rightarrow \frac{d y}{d x}=\frac{-16 x}{9 y} \end{aligned} $$
...